24

Consider the polytope in $\mathbb{R}^3$ with $8$ vertices at coordinates $(\pm 1, \pm 2, 1), (\pm 2, \pm 1, -1)$. Geometrically this looks like a cube where the top face is stretched in the direction of the $y$-axis and the bottom face is stretched in the direction of the $x$-axis, but it still has the face structure of a cube and in particular has $6$ ...

23

Your question is essentially about extension complexity. In general, the extension complexity of a polytope $P$ is the minimum number of facets over all polytopes $Q$ which project to $P$. You are interested in the extension complexity of polygons. Fiorini, Rothvoß, and Tiwary proved that regular $n$-gons have extension complexity $O(\log n)$. For lower ...

18

Note: I've decided that this answer should be rearranged a bit
so that it clearly separates the discussion of the basic properties of
the tangent bundle from the discussion of the formulae associated to a connection.
The content is the same, but I hope it's clearer.
The standard way to discuss the geometry of connections and geodesic flow 'invariantly' (...

answered Dec 5 '16 at 9:20

Robert Bryant

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16

Another, combinatorially minimal, counterexample of such a polytope $P$ (with only five facets) is the convex hull of the six vertices $(\pm2, 0, 0)$, $(\pm1, \pm1, 1)$. Its projection to the $xy$-plane is a hexagon. The minimality of the number of vertices and the number of facets is easy to prove.

15

There's a straightforward abstract answer that you may not like, but, because it clarifies your question and explains a uniform way to answer similar questions, I'll sketch it here.
First, consider a simpler problem of this kind: Suppose that one wants to describe the group of isometries of a Riemannian metric $\rho$ on a Riemannian $n$-manifold $M$. By ...

dg.differential-geometry differential-equations projective-geometry differential-operators affine-geometry

answered Jan 10 '19 at 22:09

Robert Bryant

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14

No even in dimension 1 (and multiplying the example for $\mathbb{R}$ by the small segment you get a counterexample in $\mathbb{R}^2$).
Take the set $A_n\subset \mathbb{R}$ defined as $\bigcup_{k\in \mathbb{Z}} (2k\cdot 10^{-n},(2k+1)\cdot 10^{-n})$. I claim that there exists no finite family of translates $\bigcup_{n=1}^\infty (A_n+q_n)$ which covers $[0,1]$...

10

The version you state is definitely a 20th century development, only marginally related to Von Staudt's theorem. Here is a translation of the relevant section of Karzel & Kroll's Geschichte der Geometrie seit Hilbert, p. 51 (notation should be self-explanatory):
Examples of "non-linear" collineations were given by C. Segre [Seg 1890] for ...

answered May 8 '15 at 20:01

Francois Ziegler

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9

This is possible and here is how to do this. We will use an inductive argument, assume that the statement holds for polytops of dimension $<n$ and prove it for dimension $n$.
Take any vertex $v$ of the $n$ dimensional polytop $P$ and denote by $v_1,\ldots, v_m$ all the end-points of all the edges of $P$ starting at $v$. Let $P'$ be the convex hull of $v,...

8

Bieberbach‘s 1911-12 paper (part 1, part2) proves a result about groups rather than manifolds, and it does not assume the groups to be torsion-free. In today’s language it says that a discrete, cocompact group of Euclidean isometries contains its subgroup of translations (which is necessarily a free Abelian group) as a subgroup of finite index. So you can ...

7

Here is a geometric way that turns out to be equivalent to Robert's answer (i.e., to the Klein-Grifone-Foulon approach to connections associated to a second order ordinary differential equation of a manifold).
Let $\phi_t : TM\setminus 0 \rightarrow TM\setminus 0$ be the (local) flow of the second order equation, and let $D\phi_t : T(TM\setminus 0) \...

7

The question has a negative answer if $M$ is compact. The reason for this is that any smooth function on a compact manifold has at least one critical point.
However, if you impose a weaker nondegeneracy condition, namely that the above matrix is nondegenerate only along the diagonal of $M\times M$, then the question has a positive answer and some ...

at.algebraic-topology dg.differential-geometry gt.geometric-topology smooth-manifolds affine-geometry

answered Oct 30 '14 at 11:27

Liviu Nicolaescu

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7

As Francois Ziegler pointed out, the result that you need is the so-called fundamental theorem of geometry (FTG), which states the following: if a transformation of $\mathbb R^d$ with $d\ge2$ is bijective and maps lines onto lines, then it is affine.
The main result in AMS Proc. 1999 implies that, more generally, if a transformation of $\mathbb R^d$ with $...

7

Here is a collection of examples: Let $Q:\mathbb{R}^{n+1}\to\mathbb{R}$ be a (Lorentzian) quadratic form of type $(n,1)$, and let $L^+\subset\mathbb{R}^{n+1}$ be one component of the cone of time-like vectors with respect to $Q$ (i.e., where $Q$ is negative). Set $\phi = -\log(-Q)$, which is a well defined function on $L^+$,
and let $g$ be the Hessian ...

6

WANDA SZMIELEW's article consider Tarski's axiom system for geometry, exposed into: What is elementary geometry ? (1957/58).
[Tarski] axiom system consists only of twelve short first order $\Pi \Sigma$ sentences and the continuity axiom, in case of the second order geometry, or the continuity axiom schema, in case of the first order (i.e., elementary) ...

6

The space is just $\mathbb{R} \times S^{n-1}$: each oriented hyperplane $H$ is identified by the normal unit vector $v$ together with the inner product of $v$ with an arbitrary $w \in H$.

6

In order to present affine spaces in this fashion you need a variation of operads called clones or cartesian operads (which eventually are equivalent to Lawvere theories). You can find some references on nLab.
A clone is like a symmetric operad, but with all maps between finite sets acting not just bijections. However, it is more convenient to phrase the ...

6

You don't need parallelism: a symmetric tensor of type $\binom{0}{2}$ is positive definite if and only if it is a Riemannian metric, and has signature $(1,n-1)$ (where $n$ is the dimension of your manifold) if and only if it is a Lorentzian metric, and has signature $(p,q)$ if and only if it is a $(p,q)$ pseudo-Riemannian metric. But it is also true that an ...

6

If $E$ is given as $\{ v + Aw : w \in \mathbb Q^m \}$ where $v \in \mathbb Q^n$ and $A \in M_{n, m}(\mathbb Q)$, then if $B$ is a left-inverse for $A$ we have $E \cap \mathbb Z^n \neq \varnothing \iff (1-AB)v \in (1-AB) \mathbb Z^n$. We can find a basis of this $\mathbb Z$-module and check if the coordinates of $(1-AB) v$ in that basis are integers.
For $\...

5

The intersection is only taken over halfspaces whose boundary is in $\mathfrak H$. So $\alpha_+$ is the smallest affine root which strictly contains $\alpha$. For example, if $\mathbf A=\mathbb R$ and $\mathfrak H=\mathbb Z$ then $\{\pm x+n\ge 0\}^+=\{\pm x+n+1\ge 0\}$.
The difference between affine roots in this sense and roots in the ordinary sense is ...

5

For $K=\mathbb{R}$, the positive part of your operad (mentioned in Gabriel's comment), and its algebras have been discussed by Tom Leinster and others in connection to entropy. See, for example,
https://golem.ph.utexas.edu/category/2011/05/an_operadic_introduction_to_en.html
http://www.maths.ed.ac.uk/~tl/b.pdf

answered Jan 25 '17 at 16:10

Vladimir Dotsenko

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5

Not much is known for the general case.
Let $m(k, n, q)$ denote the minimum size of an $k$-blocking set in $AG(n, q)$. Trivially we have $m(0, n, q) = q^n$ and $m(n, n, q) = 1$. By Jamison/Brouwer-Schrijver we get $m(n-1, n, q) = 1 + n(q-1)$ as you have mentioned. To at least give bounds on other values we can prove the following inequality $$qm(k, n-1, q) ...

5

Yes, this is just $\mathrm{U}(2)\ltimes \mathbf{C}^2$, that is, all such automorphisms are affine.
Indeed, let $f$ belong to your group. After composing by a translation, we can suppose that $f$ fixes zero. The tangent map of $f$ at zero preserves the Hermitian scalar product, and hence, after composing by some element of $\mathrm{U}(2)$, we can suppose that ...

5

Using the structure equations, it is not difficult to show that, if $f:(M,g)\to(N,h)$ is a diffeomorphism of (not necessarily complete) connected surfaces that is affine in the OP's sense, i.e., $\nabla(\mathrm{d}f)=0$, then $f$ has constant singular values and $L\bigl(f(p)\bigr) = K(p)/|\det(f)|$ for any $p\in M$, where $K:M\to\mathbb{R}$ and $L:N\to\mathbb{...

4

Consider a 3d-rotation with respect to the axis generated by the unit vector $(a,b,c)$ to the angle $\theta$. Its matrix is
$$
M=\pmatrix{\cos \theta+a^2(1-\cos\theta)&ab(1-\cos\theta)-c\sin\theta&
ac(1-\cos\theta)+b\sin\theta
\\ab(1-\cos\theta)+c\sin\theta&\cos \theta+b^2(1-\cos\theta)&bc(1-\cos\theta)-a\sin\theta\\ac(1-\cos\theta)-b\sin\...

4

The equality is never satisfied.
We send $T$ to $T^t$ by an affine map. This sends the unit circle to an ellipse. The bilipschitz constant is either the half of the major axis or the reciprocal of the half of the minor axis. When we look at how the sides of a triangle are distorted, we "measure the ellipse" in three directions only. There is no guarantee ...

4

The set of affinely flat structures has been studied by many authors. Among them, we can quote William Goldman who has studied the deformations of $(X,G)$-structures. Let $X$ be a manifold, $G$ a Lie group which acts on $X$, we suppose that if two elements of $G$ coincide on an open subset of $X$, they coincide on $X$. A manifold $M$ is an $(X,G)$-manifold ...

4

What you are missing is the so-called 'third fundamental (cubic) form' for this hypersurface, which, of course, is not isotropic.
Also, the 'diagonal' transformation that you write down fixes $(1,\ldots,1)$ only when it is the identity. The stabilizer of this point in the stabilizer of $C$ is actually discrete.

answered Sep 23 '14 at 17:19

Robert Bryant

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4

Each hyperplane not passing through the origin is uniquely determined by its closest point to the origin, plus sign (the normal vector points towards or away from the origin), so the set of such hyperplanes is simply $(\mathbb{R}^n \backslash \{0\})\times \{-1, 1\}.$ The two connected components are glued together along $\mathbb{S}^{n-1}.$

3

A cool generalization of Rédei's method furnishes the following lower bound:
Theorem-lower bound.
Let $U\subset \mathbb F_p$ have cardinality $2p$. Then $U$ is contained in the union of two lines or it determines at least $\lfloor\frac{p+4}{3}\rfloor$ rich directions.
The key idea is to consider the y-derivative of the Rédei polynomial
$$ H_{U}(x,y):=\...

3

By the fundamental theorem of geometry, if a transformation of $\mathbb R^d$ with $d\ge2$ is bijective and maps lines onto lines, then it is affine. The main result in AMS Proc. 1999 implies that, more generally, if a transformation of $\mathbb R^d$ with $d\ge2$ is surjective and maps lines into lines, then it is affine.
So, you don't need the condition "...

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